# Thread: Partial derivative of an integral of several variables

1. ## Partial derivative of an integral of several variables

Hi,

I'm trying to understand Fourier series and in order to do so have to be able to calculate the partial derivates of the function:
I_n=(Integral [ f(x) - a_0/2 - sigma (a_k cos(kx) + b_k sin(kx)]^2)dx
w.r.t. a_0, a_k and b_k, where the integral is calculated from -pi to pi.

I actually had certain thoughts about how to go about this, which I originally wrote here, but as I quickly realized it was completely wrong I've deleted this... Now I'm thinking like this: it's a definite integral, so if I calculate the primitive functions of all the terms containing x, and then calculate the partial derivatives that should do it, no?

Still, I don't get the answer in my book... So....

If someone could explain to me how to go about this in an easy-to-understand, step-by-step manner it'd be greatly appreciated!

/Yair

2. ## Re: Partial derivative of an integral of several variables

This is $\int_{-\pi}^{\pi} [f(x)- \frac{a_0}{2}- \sum (a_k cos(kx)+ b_k sin(kx)]^2 dx$?

The partial derivatives of $\int_a^b f(x, u, v)dx$ with respect to u and v are
$\int_a^b \frac{\partial f}{\partial u} dx$ and $\int_a^b \frac{\partial f}{\partial v}dx$.

That is, just take the partial derivatives of the integrand.

3. ## Re: Partial derivative of an integral of several variables

Originally Posted by HallsofIvy
This is $\int_{-\pi}^{\pi} [f(x)- \frac{a_0}{2}- \sum (a_k cos(kx)+ b_k sin(kx)]^2 dx$?
Yes, that's what I meant.

Originally Posted by HallsofIvy
The partial derivatives of $\int_a^b f(x, u, v)dx$ with respect to u and v are
$\int_a^b \frac{\partial f}{\partial u} dx$ and $\int_a^b \frac{\partial f}{\partial v}dx$.

That is, just take the partial derivatives of the integrand.
I see what you're saying. It'd be great if you could prove this for me!